Rewrite $\sqrt[3]{2^6\cdot3^3\cdot11^3}$ as an integer.
Answer: Starting with $2^6\cdot3^3\cdot11^3$, the cube root of that expression is $2^{6/3}\cdot3^{3/3}\cdot11^{3/3}$, which is $2^2\cdot3\cdot11=\boxed{132}$.